The partial fractions of $\frac{3\mathrm{x}-1}{\left(1-\mathrm{x}+{\mathrm{x}}^2\right)\left(2+\mathrm{x}\right)}$ are

$$\begin{gathered} \frac{3\mathrm{x}-1}{\left(1-\mathrm{x}+{\mathrm{x}}^2\right)\left(2+\mathrm{x}\right)}=\frac{\mathrm{Ax}+\mathrm{B}}{{\mathrm{x}}^2-\mathrm{x}+1}+\frac{\mathrm{C}}{\mathrm{x}+2} \\ \Rightarrow \left(3\mathrm{x}-1\right)=\left(\mathrm{Ax}+\mathrm{B}\right)\left(\mathrm{x}+2\right)+\mathrm{C}\left({\mathrm{x}}^2-\mathrm{x}+1\right) \end{gathered}$$

Comparing the coefficient of like terms, we get

$$\begin{gathered} \mathrm{A}+\mathrm{C}=0, 2\mathrm{A}+\mathrm{B}-\mathrm{C}=3, 2\mathrm{B}+\mathrm{C}=-1 \\ \Rightarrow \mathrm{A}=1, \mathrm{B}=0, \mathrm{C}=-1 \\ \therefore \frac{3\mathrm{x}-1}{\left(1-\mathrm{x}+{\mathrm{x}}^2\right)\left(2+x\right)}=\frac{\mathrm{x}}{{\mathrm{x}}^2-\mathrm{x}+1}-\frac{1}{\mathrm{x}+2}. \end{gathered}$$